I have the following theorem to prove.
Given a real number $a$, define the set $S$ such that $S = \{x \in \mathbb Q: x < a\}$. Show that $a = \sup S$.
My attempt at a proof is as follows
Proof: Let $m$ be a rational number larger than $a$. Then by the definition of the set, this element is excluded from $S$. As this element is larger than every element in $S$, this is an upper bound for $S$. Therefore $S$ has at least one upper bound. As $S$ is bounded, by the Completeness Axiom, it has a least upper bound and therefore has a supremum.
!! This is where I get stuck, I know intuitively that the number a must be the least upper bound, but I'm not exactly sure why. I know that there must be some property of the reals that is responsible for this almost obvious fact but I'm not sure what it is.
I have a feeling that this is so simple that you might accidentally answer it for me in the process of helping, but if you could offer a hint that might lead this horse to water without making it drink I would be greatly appreciative.
I would suggest a proof by contradiction. Suppose that there exists a real number $b \in \mathbb{R}$ such that $b$ is an upper bound of $\{ x \in \mathbb{Q} : x < a \}$. How do we know there exists a rational number $c$ such that $b < c < a$?